$h(t) = t^{2}+2t+2+g(t)$ $g(n) = -4n+5$ $f(t) = 3t^{2}+4t+3(g(t))$ $ f(g(2)) = {?} $
Answer: First, let's solve for the value of the inner function, $g(2)$ . Then we'll know what to plug into the outer function. $g(2) = (-4)(2)+5$ $g(2) = -3$ Now we know that $g(2) = -3$ . Let's solve for $f(g(2))$ , which is $f(-3)$ $f(-3) = 3(-3)^{2}+(4)(-3)+3(g(-3))$ To solve for the value of $f$ , we need to solve for the value of $g(-3)$ $g(-3) = (-4)(-3)+5$ $g(-3) = 17$ That means $f(-3) = 3(-3)^{2}+(4)(-3)+(3)(17)$ $f(-3) = 66$